Gas Turbine: 22-Mec-B3 Energy Conversion and Power Generation, May 2018, Q1

Question 1: Gas Turbine (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single-shaft open-cycle gas turbine draws atmospheric air at 100 kPa and 15 °C, compresses it through a pressure ratio of 12, burns 2.68 kg/s of 40 MJ/kg fuel in the air stream and expands the products back to atmosphere.

Given data (question page and page-17 constants)
Pressure ratio	$r_p = 12$
Air mass flow	$M_{\text{air}} = 142\ \text{kg/s}$
Fuel mass flow	$M_{\text{fuel}} = 2.68\ \text{kg/s}$
Fuel heating value	$\text{CV} = 40\,000\ \text{kJ/kg}$
Compressor isentropic efficiency	$\eta_c = 0.90$
Turbine isentropic efficiency	$\eta_t = 0.88$
Inlet pressure and temperature	$p_1 = 100\ \text{kPa},\ T_1 = 288.15\ \text{K}$
Cold air (page 17)	$c_p = 1.005,\ c_v = 0.718\ \text{kJ/kg K}$
Hot products (question note)	$c_p = 1.148\ \text{kJ/kg K},\ k = 1.333$

Find. The four cycle-state temperatures on a T-s diagram, the actual compressor-exit, turbine-inlet and turbine-exhaust temperatures, and the net power output and thermal efficiency of the unit.

Question 1(a) — T-s diagram of the open Brayton cycle. 1 → 2s is the isentropic compression and 1 → 2 the actual one; 2 → 3 is combustion along the 1.2 MPa isobar; 3 → 4s is the isentropic expansion and 3 → 4 the actual one; the dashed grey line 4 → 1 closes the diagram as exhaust to atmosphere. The isobars are drawn with the cold-air specific heat below the combustor and the hot-gas value above it, which is why the upper isobar is the flatter one.

Approach. Work the compressor with cold-air properties, close an energy balance on the combustor using the product mass flow and the hot-gas specific heat to obtain the turbine-inlet temperature, then expand with the hot-gas isentropic exponent and apply the two machine efficiencies.

Part (a) — identify the cycle states. The six points carried through the calculation are numbered on the diagram above: 1 at compressor inlet (100 kPa, 288.15 K), 2s and 2 at compressor exit (1.2 MPa), 3 at turbine inlet (1.2 MPa) and 4s and 4 at turbine exhaust (100 kPa). The subscript s marks the isentropic end state that each machine efficiency is measured against.
Part (b) — specific-heat ratio and isentropic exponent for cold air. From the page-17 constants, $k = c_p/c_v = 1.005/0.718 = 1.3997$, so the isentropic exponent is $$\frac{k-1}{k} = \frac{0.3997}{1.3997} = 0.28557 .$$ The compressor temperature ratio therefore follows from the isentropic relation $T_{2s}/T_1 = (p_2/p_1)^{(k-1)/k}$: $$\frac{T_{2s}}{T_1} = 12^{0.28557} = 2.0331 .$$
Isentropic and actual compressor exit temperature. Applying that ratio to the inlet temperature gives the ideal exit state, and the compressor efficiency $\eta_c = \Delta T_{\text{isentropic}}/\Delta T_{\text{actual}}$ converts it to the real one: $$T_{2s} = 288.15 \times 2.0331 = 585.9\ \text{K} = 312.7\ ^\circ\text{C},$$ $$T_2 = T_1 + \frac{T_{2s}-T_1}{\eta_c} = 288.15 + \frac{585.9-288.15}{0.90} = \boxed{618.9\ \text{K} = 345.8\ ^\circ\text{C}} .$$ The irreversibility of the machine has added 33 K to the ideal compression, which is the entropy increase shown between 2s and 2 on the diagram.
Combustor energy balance gives the turbine inlet temperature. The fuel adds its whole heating value to the stream, and the note requires the changed mass flow and specific heat to be used for the gas in the turbine, so the products flow is $M_{\text{gas}} = 142 + 2.68 = 144.68\ \text{kg/s}$ and $c_{p,\text{gas}} = 1.148\ \text{kJ/kg K}$: $$\dot{Q}_{\text{in}} = M_{\text{fuel}}\,\text{CV} = 2.68 \times 40\,000 = 107\,200\ \text{kW},$$ $$\Delta T_{\text{comb}} = \frac{\dot{Q}_{\text{in}}}{M_{\text{gas}}\,c_{p,\text{gas}}} = \frac{107\,200}{144.68 \times 1.148} = 645.4\ \text{K},$$ $$T_3 = T_2 + \Delta T_{\text{comb}} = 618.9 + 645.4 = \boxed{1264.4\ \text{K} = 991.2\ ^\circ\text{C}} .$$
Expansion through the turbine. The products expand back to 100 kPa with $k = 1.333$, so the isentropic exponent changes to $(k-1)/k = 0.333/1.333 = 0.24981$ and the turbine temperature ratio is $12^{0.24981} = 1.8603$. Solving for the ideal exhaust state and then applying $\eta_t = \Delta T_{\text{actual}}/\Delta T_{\text{isentropic}}$, $$T_{4s} = \frac{T_3}{1.8603} = \frac{1264.4}{1.8603} = 679.7\ \text{K} = 406.5\ ^\circ\text{C},$$ $$T_4 = T_3 - \eta_t (T_3 - T_{4s}) = 1264.4 - 0.88 \times 584.7 = \boxed{749.8\ \text{K} = 476.7\ ^\circ\text{C}} .$$ Note that the hot gas expands through a smaller temperature ratio than the air was compressed through (1.860 against 2.033) purely because its specific-heat ratio is lower; ignoring that would leave the exhaust more than 70 K too cold.
Part (c) — shaft powers. Each machine handles its own stream, so the turbine works on the products and the compressor on the air alone: $$P_t = M_{\text{gas}}\,c_{p,\text{gas}}(T_3-T_4) = 144.68 \times 1.148 \times 514.6 = 85\,460\ \text{kW},$$ $$P_c = M_{\text{air}}\,c_{p,\text{air}}(T_2-T_1) = 142 \times 1.005 \times 330.8 = 47\,210\ \text{kW},$$ $$P_{\text{net}} = P_t - P_c = 85\,460 - 47\,210 = \boxed{38\,260\ \text{kW} = 38.3\ \text{MW}} .$$
Thermal efficiency and back-work ratio. Dividing the net output by the fuel energy released, $$\eta = \frac{P_{\text{net}}}{\dot{Q}_{\text{in}}} = \frac{38\,260}{107\,200} = \boxed{0.357 = 35.7\ \%} .$$ The back-work ratio is $P_c/P_t = 47\,210/85\,460 = 0.552$, that is, the compressor absorbs 55 % of the gross turbine output. That is the characteristic figure for a simple-cycle machine and is the reason a gas turbine loses output so quickly when the compressor fouls or the ambient air warms.

Final results
Quantity	Symbol	Value
Compressor exit temperature	$T_2$	618.9 K = 345.8 °C
Turbine inlet temperature	$T_3$	1264.4 K = 991.2 °C
Turbine exhaust temperature	$T_4$	749.8 K = 476.7 °C
Gross turbine power	$P_t$	85.46 MW
Compressor power absorbed	$P_c$	47.21 MW
Net power output	$P_{\text{net}}$	38.3 MW
Thermal efficiency	$\eta$	35.7 %
Back-work ratio	$P_c/P_t$	0.552

Check: which specific heat closes the combustor. The note on the question page fixes the hot-gas properties for the expansion and cold-air properties for "other processes", but it also instructs the candidate to take account of the changed mass flow and specific heat when calculating the conditions of the gas in the turbine — and the turbine inlet temperature is one of those conditions. The solution above therefore closes the combustor on $M_{\text{gas}} = 144.68$ kg/s and $c_p = 1.148$. Closing it instead on the cold-air value would raise $T_3$ to 1083 °C and the output to 44.5 MW, but it also credits the cycle with energy the fuel never supplied, because the incoming air would then be carried at a different specific heat from the products it becomes. The preamble to Question 2 quotes a nominally identical machine at 42.5 MW with 2.69 kg/s of fuel and a 145 kg/s exhaust; the reading used here sits about 10 % below that figure, and either reading should be accepted provided the assumption is stated, as the paper's own Note 6 invites.

Topic. Open Brayton cycle with real machine efficiencies and variable specific heats.

Key relations. $T_{2s}/T_1 = r_p^{(k-1)/k}$; $\eta_c = (T_{2s}-T_1)/(T_2-T_1)$; $\eta_t = (T_3-T_4)/(T_3-T_{4s})$; $\dot{Q} = M c_p \Delta T$; $\eta = (P_t - P_c)/\dot{Q}_{\text{in}}$.

Why this works. A gas turbine is an open system, so the working fluid is air in the compressor and combustion products in the turbine. Those products are heavier in triatomic molecules, which raises the specific heat to about 1.148 kJ/kg K and lowers the specific-heat ratio to about 1.333. Because the isentropic temperature ratio is governed by $(k-1)/k$, the same pressure ratio delivers a smaller temperature ratio in the turbine than in the compressor. That is favourable: the turbine expands a larger mass through a smaller temperature ratio but at a higher specific heat, and the product of the three is what buys the net work. The machine efficiencies enter as temperature-difference ratios rather than enthalpy ratios only because constant specific heats have been assumed on each side.

Common pitfall. Using one specific heat and one mass flow throughout. Doing so understates the turbine work by roughly 12 % and moves the efficiency by several points; equally common is inverting one of the two efficiency definitions, which for the compressor gives a temperature below the isentropic value and should be caught immediately by inspection.

Source. Çengel & Boles, Thermodynamics 9th ed., §9.8–9.9 (Brayton cycle, deviation of actual cycles); El-Wakil, Powerplant Technology, Ch. 8.

Question 1 of 8: Gas Turbine

Question 1: Gas Turbine (10 marks)